ISRO CSE 2023 | Question: 71

Question

The time delay obtained through an $8$-bit serial register with $400$ MHz clock is:

1. $20$ ns
2. $2.5$ µs
3. $20$ µs
4. $2.5$ µs

Answer 1

Determine the clock period:

• Clock period (T) = 1 / Clock frequency
• T = 1 / 400 MHz = 1 / (400 * 10^6) seconds = 2.5 * 10^-9 seconds = 2.5 ns

 Calculate the time delay:

• Time delay = Clock period * Number of bits
• Time delay = 2.5 ns/bit * 8 bits = 20 ns

Therefore, the time delay obtained through an 8-bit serial register with a 400 MHz clock is 20 ns.

So, the correct answer is (a) 20 ns